Question

The value of the integral $\underset{\frac{\pi }{6}}{\overset{\frac{\pi }{2}}{\int }}\cos \left(\frac{1}{\sin \left(\frac{1}{\sin x}\right)}\right)\cdot \cos \left(\frac{1}{\sin x}\right)\cdot \cos x\cdot \frac{1}{{\sin }^{2}x\cdot {\sin }^2\left(\frac{1}{\sin x}\right)}dx$ equals

• A. $\sin (\mathrm{cosec}1)-\sin (\mathrm{cosec}2)$
• B. $\cos (\mathrm{cosec}1)-\cos (\mathrm{cosec}2)$
• C. $2(\cos (\mathrm{cosec}1)-\cos (\mathrm{cosec}2))$
• D. $\sin (\sin 2)-\sin (\sin 1)$

Answer 1

Answer: (A)

$\text{ Consider }\frac{d}{dx}\sin \cdot \frac{1}{\sin \left(\frac{1}{\sin x}\right)}$
$\cos \frac{1}{\sin \left(\frac{1}{\sin x}\right)}\cdot (-)\frac{1}{{\sin }^2\left(\frac{1}{\sin x}\right)}\cdot \cos \left(\frac{1}{\sin x}\right)\cdot (-1)\frac{1}{{\sin }^{2}x}\cdot \cos x$
$=\cos \frac{1}{\sin \left(\frac{1}{\sin x}\right)}\cdot \cos \left(\frac{1}{\sin x}\right)\cdot \cos x\cdot \frac{1}{{\sin }^{2}x\cdot {\sin }^2\left(\frac{1}{\sin x}\right)}$
hence the given integrand is $\frac{d}{dx}\sin \frac{1}{\sin \left(\frac{1}{\sin x}\right)}$
$\therefore I=\underset{\frac{\pi }{6}}{\overset{\frac{\pi }{3}}{\int }}\left(\frac{d}{dx}\left(\sin \frac{1}{\sin \left(\frac{1}{\sin x}\right)}\right)\right)dx={\left(\sin \frac{1}{\sin \left(\frac{1}{\sin x}\right)}\right)}_{\pi /6}^{\pi /2}=\sin (\mathrm{cosec}1)-\sin (\mathrm{cosec}2)$